Problem: An equilateral triangle has an area of $64\sqrt{3}$ $\text{cm}^2$. If each side of the triangle is decreased by 4 cm, by how many square centimeters is the area decreased?
Explanation: We first consider an equilateral triangle with side length $s$. If we construct an altitude, it will divide the equilateral triangle into two congruent $30-60-90$ triangles with the longest side having length $s$ and the altitude opposite the $60^\circ$ angle. Since the side lengths of a $30-60-90$ triangle are in a $1:\sqrt{3}:2$ ratio, the altitude will have length $\frac{s\sqrt{3}}{2}$. Since the base of this equilateral triangle is $s$, its area will be $\frac{1}{2}{b}{h}=\frac{1}{2}s \left(\frac{s\sqrt{3}}{2}\right)=\frac{s^2 \sqrt{3}}{4}$.

Now we can set this expression equal to $64\sqrt{3}$ and solve for $s$ to find the side length of our original triangle. Doing this, we get that $\frac{s^2 \sqrt{3}}{4}=64\sqrt{3}$. We can then multiply both sides of the equation by $\frac{4}{\sqrt{3}}$ to get that $s^2=256$. Taking the square root of both sides, we find that $s=16$, so the original triangle had a side length of $16$ cm. If we decrease this by $4$ cm, we get that the new triangle has side length $12$ cm and therefore has an area of $\frac{144 \sqrt{3}}{4}=36\sqrt{3}$ cm. Therefore, the area is decreased by $64\sqrt{3}-36\sqrt{3}=\boxed{28\sqrt{3}}$ cm.